Question

28. An RC circuit is connected to a 12.0 V battery at t=0 s and the capacitor voltage is monitored as a function of time. This cap is charging! a) Att = 0.147 s, the capacitor voltage is 20.0% of the battery voltage. What is the RC time constant? b) When will the capacitor be charged to 99.0% of the battery voltage? c) If the cap has value 2.0 mF, what is the resistance? d) After the time of part (b), the cap is allowed to discharge through the resistor. i) What is the current through the resistor after 0.448 s? ii) What is the voltage across the resistor at 0.448 s? 111) What is the power dissipated in the resistor at 0.448 s? iv) How much energy is stored in the cap at 0.448 s?

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Answer 1

A VCE +12V The voltage across the chacitor at any time to in given by [ &(t) = v (1-eleco (where RC = C = time constant) (a) given at t=0 11475: u(t) = 20% of v. =) V(t) = 0.2 x V - V.lt) - 0.2 from eqr. : at t= 0.147, so Velt). 0.2 = (1 - e

=> 02: (i- etli) -) & bc - 1-0.2 = 0.8 - - RC = ln (0.8) (where 1s()) is log bage é ) =) RC = -t = 0.07069 In (O. -) RC = 0.0707s 6 when u(t) = 0.99 V: => -> Vict) - 0.99 = (1-e thre) v & the = 0.01 -tlec - in 0.01) ti - RC , in (0.0!) - - 0.0707 x ln (0.01) – 0.32 56 s (1) Given (= 2 ut. = RC = 0.0707 s. -) RE 0.07075 - 35 35 en 2nF

(d) after t:o 32565 volt) 0.ga v.0.49x120 -0. 1188V During Discharging of cofactor Velt) = Vetlac (1) at t= 0 4485 So, time geh between discharging t'0.4485 - 0:3265 = 0.122s So, V. (t)= vet/re home, vollege ocross resistor is: V VE V- ¥ (") - vi-e-t/a) Va(t) = 12v ( 1 & 0122.07075) • 12 V X 0.823 - 9.878 v IR (current through resistor) > Te(t) = Vect) - 92 9.878 V 35.35 R - 0.279 A (1) Voltage across e(t) = Resistor at tori8s & 9.878 v

;) Power dissipted P= VI - VR (+) IR (A) - 9.878 V & 0.279 A - 2.756 W Energy stored (u) in Cohacitor U= cv? at t= 0.4485 where V = Velt) = V-Vp (t) - 12v- 9.878V = 2.122 v =) = 2 x 10306 x (2.12203 & 4.503 +10 -3 J

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